L(u)=0, L=d/dt-a-D*Laplacian.
Here, L is called a differential operator that works on the function u.

To show that L is linear, you must show that for any functions U,u , and constants c, b that:
L(c*U+b*u)=c*L(U)+b*L(u)

We have:
L(c*U+b*u)=d/dt(c*U+b*u)-a*(c*U+b*u)+D*Laplacian(c*U+b*u)=
=c*(dU/dt-a*U+D*laplacian(U))+b*(dU/dt-a*U+D*laplacian(U))=
c*L(U)+b*L(u),
as required.
Note that if U and u are both solutions to the original differential equation;
i.e. L(U)=L(u)=0,
the linearity of L implies that any sum of U and u also is a solution.